exam:Back:Rn:FD:xyz: Employing defn:CCS:Bregman:leftdefn:CCS:Bregman:left:, [1]} and eq:exam:back:R:negative:xyz:D, we observe that in this case, \( p \in \overleftarrow{CC{}}_{f}(\mathcal {S}) \)  if and only if \(p=x + \alpha (y-x) +\beta (z-x) \in \left[0, 1\right]^{n}\)  such that
\(&D_{f} (p,x) =D_{f} (p,y)=D_{f} (p,z) \\\Leftrightarrow &{\left\lbrace \begin{array}{ll}\sum ^{n}_{i=1} p_{i} \ln (\frac{p_{i}}{x_{i}}) + (1 -p_{i})\ln (\frac{1-p_{i}}{1-x_{i}}) = \sum ^{n}_{i=1} p_{i} \ln (\frac{p_{i}}{y_{i}}) + (1 -p_{i})\ln (\frac{1-p_{i}}{1-y_{i}}) \\\sum ^{n}_{i=1} p_{i} \ln (\frac{p_{i}}{x_{i}}) + (1 -p_{i})\ln (\frac{1-p_{i}}{1-x_{i}}) = \sum ^{n}_{i=1} p_{i} \ln (\frac{p_{i}}{z_{i}}) + (1 -p_{i})\ln (\frac{1-p_{i}}{1-z_{i}})\end{array}\right.}\\\Leftrightarrow &{\left\lbrace \begin{array}{ll}\sum ^{n}_{i=1} p_{i} \ln \left( \frac{x_{i}}{y_{i}} \frac{1-y_{i}}{1-x_{i}} \right) = \sum ^{n}_{i=1} \ln \left( \frac{1-y_{i}}{1-x_{i}} \right) \\\sum ^{n}_{i=1} p_{i} \ln \left( \frac{x_{i}}{z_{i}} \frac{1-z_{i}}{1-x_{i}} \right) = \sum ^{n}_{i=1} \ln \left( \frac{1-z_{i}}{1-x_{i}} \right)\end{array}\right.}\\\Leftrightarrow &{\left\lbrace \begin{array}{ll}\alpha \sum ^{n}_{i=1} (y_{i} -x_{i} ) \ln \left( \frac{x_{i}}{y_{i}} \frac{1-y_{i}}{1-x_{i}} \right) +\beta \sum ^{n}_{i=1} (z_{i} -x_{i} ) \ln \left( \frac{x_{i}}{y_{i}} \frac{1-y_{i}}{1-x_{i}} \right)= \sum ^{n}_{i=1} \ln \left( \frac{1-y_{i}}{1-x_{i}} \right) -x_{i} \ln \left( \frac{x_{i}}{y_{i}} \frac{1-y_{i}}{1-x_{i}} \right)\\\alpha \sum ^{n}_{i=1} (y_{i} -x_{i} ) \ln \left( \frac{x_{i}}{z_{i}} \frac{1-z_{i}}{1-x_{i}} \right) +\beta \sum ^{n}_{i=1} (z_{i} -x_{i} ) \ln \left( \frac{x_{i}}{z_{i}} \frac{1-z_{i}}{1-x_{i}} \right)= \sum ^{n}_{i=1} \ln \left( \frac{1-z_{i}}{1-x_{i}} \right) -x_{i} \ln \left( \frac{x_{i}}{z_{i}} \frac{1-z_{i}}{1-x_{i}} \right)\end{array}\right.},\) 
